Use the definition of the limit to prove the given limit.

**Solution**

Let ε> 0 is any number then we have to find a number δ > 0 so that the following will be true.

| √x - 0| < ε whenever 0 < x - 0 < δ

Or upon a little simplification we need to show,

√x < ε whenever 0 < x < δ

Let's begin with the left hand inequality and illustrates if we can't utilizes that to get a guess for δ . The only simplification which we really have to do here is to square both sides.

√x < ε whenever x < δ^{2}

Hence, it looks like we can chose δ = ε ^{2} .

Let's verify this. Let ε > 0 be any number and select δ = ε ^{2} .. Next suppose that 0 < x < ε ^{2}. it gives,

|√x -0|= √x some quick simplification

< (√ ε)^{2} use the assumption that |x| < ε

= ε simplify

Now we illustrated that,

|√x - 0| = √x whenever 0 < x - 0 < ε ^{2}

and therefore by the definition of the right-hand limit we have,